Two-Dimensional Elastic Collision of Equal Masses

  • #1

Homework Statement


Show that if an elastic collision between a mass and a stationary target of equal mass is not head-on that the projectile and target final velocities are perpendicular. (Hint: Square the conservation of momentum equation, using ##p^2=p\cdot p##, and compare the resulting equation with the energy conservation equation.)

Homework Equations


##p=mv##
##K=\frac{1}{2}mv^2##

The Attempt at a Solution


So I followed the hint and got $$\overrightarrow{p_1^2}=\overrightarrow{p_1^{'2}}+\overrightarrow{p_2^{'2}}+2\left(\overrightarrow{p_1}\cdot \overrightarrow{p_2}\right)$$ Plugging in values I got $$m^2\overrightarrow{v_1^2}=m^2\overrightarrow{v_1^{'2}}+m^2\overrightarrow{v_2^{'2}}+2\left(m\overrightarrow{v}_1\cdot m\overrightarrow{v}_2\right)$$ $$\overrightarrow{v_1^2}=\overrightarrow{v_1^{'2}}+\overrightarrow{v_2^{'2}}$$ For the x- and y-components I got the following assuming the target traveled along the x-axis after the collision $$v_1^2cos^2(\theta)=v_1^{'2}cos^2(\theta ^{'})+v_2^{'2}$$ $$v_1^2sin^2(\theta)=v_1^{'2}sin^2(\theta ^{'})$$ This is where I got stuck and wasn't sure how to solve for ##v_1^{'}## or ##v_2^{'}##. I'm not entirely sure when or how to apply the equation for kinetic energy. Am I in the right direction or did I do something wrong?
 
Last edited:

Answers and Replies

  • #2
kuruman
Science Advisor
Homework Helper
Insights Author
Gold Member
10,265
3,392
You did something wrong. The last two equations are dimensionally incorrect. Review their derivation.
 
  • #3
You did something wrong. The last two equations are dimensionally incorrect. Review their derivation.
I forgot to square the initial velocity. Were there any other mistakes?
 
  • #4
I think I got it.

So the momentum in the x and y-direction is $$v_1=v_1^{'}cos(\theta _1)+v_2^{'}cos(\theta _2)$$ $$0=v_1^{'}sin(\theta _1)-v_2^{'}sin(\theta _2)$$ I then used the following to find the magnitude of the momentum $$p^2=p_x^2+p_y^2$$ Doing this I got $$v_1=v_1^{'2}(cos^2(\theta _1)+sin^2(\theta _1))+v_2^{'2}(cos^2(\theta _2)+sin^2(\theta _2))+2v_1^{'}v_2^{'}(cos(\theta _1)cos(\theta _2)-sin(\theta _1)sin(\theta _2))$$ Using trig identities and the law of cosines, I simplified it to $$v_1^2=v_1^{'2}+v_2^{'2}+2v_1^{'}v_2^{'}cos(\theta _1+\theta _2)$$ Knowing that kinetic energy is conserved in an elastic collision I used the equation for kinetic energy and simplified $$\frac{1}{2}mv_1^2=\frac{1}{2}mv_1^{'2}+\frac{1}{2}mv_2^{'2}$$ $$v_1^2=v_1^{'2}+v_2^{'2}$$ With this formula, ##v_1^2## can be substituted for ##v_1^{'2}+v_2^{'2}## After substituting the equation simplifies to $$0=2v_1^{'}v_2^{'}cos(\theta _1+\theta _2)$$ Solving results in $$v_1^{'}=0$$ $$v_2^{'}=0$$ $$cos(\theta _1+\theta _2)=0$$ Solving for the last equation results in $$\theta _1+\theta _2 = 90°\text{ or }\frac{\pi}{2}$$ Thus, the two velocities must be perpendicular.
 
  • #5
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
36,231
6,842
Solving results in $$v_1^{'}=0$$ $$v_2^{'}=0$$ $$cos(\theta _1+\theta _2)=0$$
it results in at least one of those three. How do you rule out the first two?
 
  • #6
it results in at least one of those three. How do you rule out the first two?
The question asks to prove that the two velocities will be perpendicular, so I assumed the velocities to be non-zero because otherwise there would be no answer as the angle between the two velocities would not exist.
 
  • #7
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
36,231
6,842
The question asks to prove that the two velocities will be perpendicular, so I assumed the velocities to be non-zero because otherwise there would be no answer as the angle between the two velocities would not exist.
You are given that the collision is not head-on. From that you can show neither velocity is zero.
 
  • #8
You are given that the collision is not head-on. From that you can show neither velocity is zero.
That makes sense, thank you very much.
 
  • #9
kuruman
Science Advisor
Homework Helper
Insights Author
Gold Member
10,265
3,392
Here is a quick geometric proof, not that your algebraic proof is incorrect.
Momentum conservation for a two equal-mass collision requires that ##\vec{v}_1={\vec{v_1}'}+{\vec{v_2}'}##.
Since the collision is not one-dimensional, if you draw a vector addition diagram, you get a closed triangle with the magnitudes of the three vectors as its sides.
Energy conservation for a two equal-mass collision requires that ##v_1^2= {v'_1}^2+{v'_1}^2##.
Therefore, said triangle obeys the Pythagorean theorem with ##v_1## as the hypotenuse and ##{v'_1}## and ##{v'_2}## as the two right sides.
 
  • #10
Here is a quick geometric proof, not that your algebraic proof is incorrect.
Momentum conservation for a two equal-mass collision requires that ##\vec{v}_1={\vec{v_1}'}+{\vec{v_2}'}##.
Since the collision is not one-dimensional, if you draw a vector addition diagram, you get a closed triangle with the magnitudes of the three vectors as its sides.
Energy conservation for a two equal-mass collision requires that ##v_1^2= {v'_1}^2+{v'_1}^2##.
Therefore, said triangle obeys the Pythagorean theorem with ##v_1## as the hypotenuse and ##{v'_1}## and ##{v'_2}## as the two right sides.
Thank you! I wish I had seen that sooner.
 

Related Threads on Two-Dimensional Elastic Collision of Equal Masses

Replies
1
Views
354
Replies
15
Views
5K
Replies
6
Views
15K
  • Last Post
Replies
2
Views
3K
Replies
3
Views
11K
Replies
2
Views
4K
Replies
1
Views
1K
Replies
2
Views
5K
  • Last Post
Replies
2
Views
657
  • Last Post
Replies
1
Views
8K
Top